Question

You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 1515 lb and was traveling eastward. Car B weighs 1125 lb and was traveling westward at 42.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 17.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750 . How fast (in miles per hour) was car A traveling just before the collision?

Answer 1

Answer:

29.4 m/s

Explanation:

mass A (Ma) = 1515 lb = 687.2 kg

mass B (Mb) = 1125 lb = 510.3 kg

distance (d) = 17.5 ft = 5.33 m

velocity B (Vb) = 42 mph = 18.78 m/s

velocity A (Va) = ?

coefficient of kinetic friction (μk) = 0.75

acceleration due to gravity (g) = 9/8 m/s

after collision the kinetic energy of the system went into the work against friction force, therefore

work done by frictional force = kinetic energy

μk x ( Ma + Mb ) x g x d = 1/2 x ( Ma + Mb ) x ( Ma + Mb ) x Vf^2

Vf = $\sqrt{2Ugd}$

from the law of conservation of momentum we have

MaVa - MbVb = (Ma + Mb) x Vf

substituting Vf = $\sqrt{2Ugd}$ into the equation above we have

 MaVa - MbVb = (Ma + Mb) x $\sqrt{2Ugd}$

Va = (((Ma + Mb) x $\sqrt{2Ugd}$) + MbVb) / Ma

Va =  (((687.2 + 510.3 ) x $\sqrt{2 x 0.75 x 9.8 x 5.33}$) + 510.3 x 18.78) / Ma

= 29.4 m/s

Answer 2

Answer:

• Car A was travelling at $34.43mph$ just before the collision

Explanation:

Let combined speed of the collision = V

therefore, by energy theorem

$\frac{1}{2}mV^2 = umgd\\\\V = \sqrt{2ugd}\\\\V = \sqrt{2*0.75*32*17.5}\\\\V = 28.983 = 19.76mph$

Let speed of A before collision = $V_A$

$1515V_A - 1125*42 = (1515+1125)*19.76\\\\1515V_A = 52166.4\\\\V_A = 34.43mph$

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